If $G$ is open and dense subset in $\mathbb R$ then show that $G\setminus\{x\}$ is also open and dense in $\mathbb R$. is it true in general metric space?
I know as $G$ is open and singleton set $\{x\}$ is closed so that $G\setminus\{x\}$ has to be open, but how to show it is dense in $\mathbb R$. next the same will true in general Metric space.
On
It's not true in a general metric space. Let $X$ be a metric space consisting of a single point $x$, with define $d(x,x) = 0$. Then $G = \{x\}$ is open and dense in $X$, but $G \setminus \{x\} = \emptyset$ is no longer dense.
On
First, we prove that $G-\{x\}$ is open and dense in general topological space $X$ of $T_2$ (Hausdorff space) and above.
Since $G-\{x\}=G\cap \{x\}^c$, and $\{x\}^c$ is open ($\{x\}$ is closed), $G-\{x\}$ is open.
Next we prove that $(G-\{x\})^c$ is nowhere dense in $X$. Since $$ (G-\{x\})^c=(G\cap \{x\}^c)^c=G^c\cup \{x\} $$ if $\varnothing\ne O\subset(G-\{x\})^c$, $O$ is open, then $O\subset G^c$ for $\{x\}$ is closed, and so $O\subset (G^c)^o$. Since $G$ is dense in $X$, $G^c$ is nowhere dense in $X$, i.e. $(G^c)^o=\varnothing$. But this means that $O=\varnothing$, contradiction. Thus $(G-\{x\})^c$ is nowhere dense in $X$, and $(G-\{x\})$ is dense in $X$.
Edit: Since $\Bbb{R}$ is Hausdorff space, above hold in $\Bbb{R}$. For general metric space that $X$ has isolated point, above doesn't hold.
Take as your universe an interval and an isolated point. As universe it is open and dense. Remove the isolated point, what remains is open (as its complement is the closed point) but sadly not dense anymore.
For your question you would need the property that every point is a density point of the universe (or the dense set).